Theory and Problems of BEGINNING CHEMISTRY Third Edition phần 1 pps

[...]... Chapter 17 CONTENTS Acid-Base Theory 17 .1 17.2 17 .3 17 .4 17 .5 17 .6 Chapter 18 Introduction The Brønsted-Lowry Theory Acid-Base Equilibrium Autoionization of Water The pH Scale Buffer Solutions Organic Chemistry 18 .1 18.2 18 .3 18 .4 18 .5 18 .6 18 .7 18 .8 18 .9 18 .10 18 .11 18 .12 Chapter 19 Nuclear Reactions 19 .1 19.2 19 .3 19 .4 19 .5 19 .6 APPENDIX Introduction Bonding in Organic Compounds Structural, Condensed, and. .. mega kilo deci centi milli micro nano pico G M k d c m µ n p 1 000 000 000 1 000 000 10 00 0 .1 0. 01 0.0 01 0.000 0 01 1 × 10 −9 1 × 10 12 1 Gm = 1 000 000 000 m 1 Mm = 1 000 000 m 1 km = 10 00 m 1 dm = 0 .1 m 1 cm = 0. 01 m 1 mm = 0.0 01 m 1 µm = 0.000 0 01 m 1 nm = 1 × 10 −9 m 1 pm = 1 × 10 12 m EXAMPLE 2.4 Since meter is abbreviated m (Table 2 -1) and milli is abbreviated m (Table 2-2), how can you tell the... coefficient, a coefficient of 1 is implied: 1 × 10 × 10 × 10 × 10 × 10 = 10 0 000 There are five 10 s multiplying the implied 1 EXAMPLE 2 .15 What is the value of 10 1 ? of 10 0 ? Ans 10 1 = 10 There is one 10 multiplying the implied coefficient of 1 100 = 1 There are no 10 s multiplying the implied coefficient of 1 EXAMPLE 2 .16 Write 2.0 × 10 4 in decimal form Ans 2.0 × 10 × 10 × 10 × 10 = 20 000 When scientists... We know that 10 dimes = 1 dollar 1 dime = 0 .1 dollar or We may divide both sides of the first of these equations by 10 dimes or by 1 dollar, yielding 10 dimes 1 dollar = 10 dimes 10 dimes 10 dimes 1 dollar = 1 dollar 1 dollar or Since the numerator and denominator (top and bottom) of the fraction on the left side of the first equation are the same, the ratio is equal to 1 The ratio 1 dollar /10 dimes is... method, we find 10 .00 mL 19 .3 g 1 mL = 19 3 g (The density is given to three significant digits.) With the equation: d = m V m = V d = (10 .00 mL) 19 .3 g 1 mL = 19 3 g EXAMPLE 2.33 What is the volume of 72.4 g of lead? (Density = 11 .3 g/mL.) Ans V = or m 72.4 g = = 6. 41 mL d 11 .3 g/mL 72.4 g 1 mL 11 .3 g = 6. 41 mL CHAP 2] 23 MATHEMATICAL METHODS IN CHEMISTRY Use of the equation requires manipulation of the basic... between them Often it is necessary to multiply by a factor raised to a power Consider the problem of changing 5.00 m3 to cubic centimeters: 5.00 m3 10 0 cm 1m If we multiply by the ratio of 10 0 cm to 1 m, we will still be left with m2 (and cm) in our answer We must multiply by (10 0 cm/m) three times; 5.00 m3 10 0 cm 1m 10 0 cm 1m 3 3 = 5 000 000 cm3 10 0 cm 1m means 10 0 cm 1m 10 0 cm 1m and includes 10 03 cm3... cm3 in the numerator and 1 m3 in the denominator EXAMPLE 2 .12 How many liters are there in 1 m3 ? Ans 1 m3 10 dm 1m 3 1L 1 dm3 = 10 00 L One cubic meter is 10 00 L The liter can have prefixes just as any other unit can Thus 1 mL is 0.0 01 L, and 1 kL is 10 00 L = 1 m3 Mass Mass is a measure of the quantity of material in a sample We can measure that mass by its weight—the attraction of the sample to the... the speed in feet per second of a jogger running 7.50 miles per hour (mi/h) 7.50 mi 5280 ft 1h 1 mi = 39 600 ft 1h 1h 39 600 ft 1h 60 min = 660 ft 1 min 660 ft 1 min 1 min 60 s Ans = 11 .0 ft 1s Alternatively, 7.50 mi 5280 ft 1h 1 mi 1h 60 min 1 min 60 s = 11 .0 ft 1s It is usually more reassuring, at least at the beginning, to do such a problem one step at a time But if you look at the combined solution,... 1 meter 2.54 centimeters 1 liter 1 kilogram 28.35 grams 39.37 inches 1 inch 1. 06 U.S quarts 2.2045 pounds (avoirdupois) 1 ounce 14 MATHEMATICAL METHODS IN CHEMISTRY [CHAP 2 EXAMPLE 2.5 (a) How many feet (ft) are there in 1. 450 miles (mi)? (b) How many meters (m) are there in 1. 450 kilometers (km)? (a) 1. 450 mi 5280 ft 1 mi (b) 1. 450 km Ans 10 00 m 1 km = 7656 ft = 14 50 m You can do the calculation of. .. that the coefficient has one and only one digit to the left of the decimal point, and that digit is not zero That notation is called standard exponential form, or scientific notation CHAP 2] 17 MATHEMATICAL METHODS IN CHEMISTRY EXAMPLE 2 .17 Write 455 000 in standard exponential form 455 000 = 4.55 × 10 × 10 × 10 × 10 × 10 = 4.55 × 10 5 Ans The number of 10 s is the number of places in 455 000 that the . Equations 14 7 10 .5 Heat Capacity and Heat of Reaction 14 7 CONTENTS ix Chapter 11 Molarity 16 2 11 .1 Introduction 16 2 11 .2 Molarity Calculations 16 2 11 .3 Titration 16 4 11 .4 Stoichiometry in Solution 16 6 Chapter. 16 6 Chapter 12 Gases 17 3 12 .1 Introduction 17 3 12 .2 Pressure of Gases 17 3 12 .3 Boyle’s Law 17 4 12 .4 Graphical Representation of Data 17 5 12 .5 Charles’ Law 17 7 12 .6 The Combined Gas Law 18 0 12 .7 The. Gas Law 18 1 12 .8 Dalton’s Law of Partial Pressures 18 3 Chapter 13 Kinetic Molecular Theory 19 5 13 .1 Introduction 19 5 13 .2 Postulates of the Kinetic Molecular Theory 19 5 13 .3 Explanation of Gas